Given that $S, T \in \mathcal{L}(V)$, where $V$ is a vector space, and that the range of $S \subseteq \text{null}(T)$, we need to prove that $(ST)^2 = 0$, i.e., $STST = 0$.

Step-by-step proof:

Step 1: Understand the given condition

• $S$ and $T$ are linear operators on the vector space $V$, meaning $S, T: V \to V$.
• The range of $S$, denoted $\text{range}(S)$, is contained within the null space of $T$, i.e., $\text{range}(S) \subseteq \text{null}(T),$ which means for every vector $v \in V$, if $v = S(w)$ for some $w \in V$, then $T(S(w)) = 0$. In other words: $T(S(w)) = 0 \text{ for all } w \in V.$

Step 2: Apply the given condition to the composition of operators

Let $v \in V$ be an arbitrary vector. We need to show that $(ST)(ST)(v) = 0$. To do so, first compute $ST(v)$ and then apply $ST$ again.

1. First application of $ST$:

   • Apply $T$ to $v$, which results in $T(v) \in V$.
   • Then apply $S$ to $T(v)$, so we get: $ST(v) = S(T(v)).$ This vector, $S(T(v))$, belongs to $\text{range}(S)$, and by the given condition, we know that: $T(S(T(v))) = 0.$

2. Second application of $ST$:

   • Now apply $ST$ again to $ST(v)$. That is, apply $ST$ to $S(T(v))$: $ST(ST(v)) = S(T(S(T(v)))).$ Since $T(S(T(v))) = 0$, this simplifies to: $S(0) = 0.$

Thus, for any $v \in V$, we have $ST(ST(v)) = 0$. Therefore, $(ST)^2 = 0$, or equivalently $STST = 0$, as required.

Conclusion:

We have shown that $(ST)^2 = 0$ by leveraging the fact that $\text{range}(S) \subseteq \text{null}(T)$. This completes the proof.

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Would you like further clarification or more details on any part of the proof?

Here are some related questions to explore:

1. What is the significance of the range and null space in linear transformations?
2. How would the proof change if the condition were $\text{null}(S) \subseteq \text{range}(T)$?
3. Can we generalize this result for higher powers, such as $(ST)^k = 0$ for $k > 2$?
4. What are some examples of linear transformations where this condition holds?
5. How does this relate to the concept of nilpotent operators?

Tip: In linear algebra, understanding how the range and null space of operators interact can often reveal important structural properties about the operators, such as being nilpotent.